Advanced Placement
Program
AP® Physics C: Mechanics
Practice Exam
The questions contained in this AP® Physics C: Mechanics Practice Exam are written to the content
specifications of AP Exams for this subject. Taking this practice exam should provide students with an
idea of their general areas of strengths and weaknesses in preparing for the actual AP Exam. Because this
AP Physics C: Mechanics Practice Exam has never been administered as an operational AP Exam,
statistical data are not available for calculating potential raw scores or conversions into AP grades.
This AP Physics C: Mechanics Practice Exam is provided by the College Board for AP Exam preparation.
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Contents
Directions for Administration ............................................................................................ ii
Section I: Multiple-Choice Questions ................................................................................ 1
Section II: Free-Response Questions .............................................................................. 14
Student Answer Sheet for Multiple-Choice Section ...................................................... 22
Multiple-Choice Answer Key........................................................................................... 23
Free-Response Scoring Guidelines.................................................................................. 24
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-i-
AP® Physics C: Mechanics
Directions for Administration
The AP Physics C: Mechanics Exam is one and one-half hours in length and consists of a multiple-choice section
and a free-response section.
•
The 45-minute multiple-choice section contains 35 questions and accounts for 50 percent of the final
grade.
•
The 45-minute free-response section contains 3 questions and accounts for 50 percent of the final grade.
Students should be given a 10-minute warning prior to the end of each section of the exam. A 10-minute break
should be provided after Mechanics is completed if students are taking Physics C: Electricity and Magnetism
immediately after Mechanics.
The actual AP Physics C Exams are administered in one session, Mechanics first followed by Electricity and
Magnetism. Students taking only one of the exams will have the most realistic experience if both sections are
completed in one session. Similarly, students taking both Physics C exams will have the most realistic experience
if both exams are completed in one session and a complete morning or afternoon is available to administer them.
If a schedule does not permit one time period for administration, it would be acceptable to administer Mechanics
on one day and Electricity and Magnetism on a subsequent day, or to further break things up and administer
Section I and Section II of each exam on subsequent days.
Many students wonder whether or not to guess the answers to the multiple-choice questions about which they are
not certain. It is improbable that mere guessing will improve a score. However, if a student has some knowledge
of the question and is able to eliminate one or more answer choices as wrong, it may be to the student’s advantage
to answer such a question.
•
The use of calculators is permitted only on Section II. Straightedges or rulers are allowed on both parts of
the exam.
•
It is suggested that the practice exam be completed using a pencil to simulate an actual administration.
•
Teachers will need to provide paper for the students to write their free-response answers. Teachers should
provide directions to the students indicating how they wish the responses to be labeled so the teacher will
be able to associate the student’s response with the question the student intended to answer.
•
The 2008–2009 AP Physics C table of information is included as a part of Section I. The table and the AP
Physics C equation lists are included with Section II. The equation lists are not allowed for Section I. If
you use these exams in subsequent years you should download the newer versions of the table and lists
from AP Central.
•
Remember that students are not allowed to remove any materials, including scratch work, from the testing
site.
-ii-
Section I
Multiple-Choice Questions
-1-
TABLE OF INFORMATION FOR 2008 and 2009
CONSTANTS AND CONVERSION FACTORS
Proton mass, m p = 1.67 ¥ 10 -27 kg
Neutron mass, mn = 1.67 ¥ 10 -27 kg
Electron mass, me = 9.11 ¥ 10 -31 kg
Avogadro’s number, N 0 = 6.02 ¥ 10 23 mol -1
Universal gas constant,
R = 8.31 J (mol iK)
e = 1.60 ¥ 10 -19 C
Electron charge magnitude,
1 electron volt, 1 eV = 1.60 ¥ 10 -19 J
Speed of light,
Universal gravitational
constant,
Acceleration due to gravity
at Earth’s surface,
c = 3.00 ¥ 108 m s
G = 6.67 ¥ 10 -11 m 3 kgis2
g = 9.8 m s2
Boltzmann’s constant, k B = 1.38 ¥ 10 -23 J K
1 u = 1.66 ¥ 10 -27 kg = 931 MeV c 2
1 unified atomic mass unit,
h = 6.63 ¥ 10 -34 J is = 4.14 ¥ 10 -15 eV is
Planck’s constant,
hc = 1.99 ¥ 10 -25 J im = 1.24 ¥ 103 eV i nm
⑀0 = 8.85 ¥ 10 -12 C2 N im 2
Vacuum permittivity,
Coulomb’s law constant, k = 1 4 p⑀ 0 = 9.0 ¥ 109 N im 2 C2
m0 = 4 p ¥ 10 -7 (T im) A
Vacuum permeability,
Magnetic constant, k ¢ = m0 4 p = 10 -7 (T im) A
1 atm = 1.0 ¥ 105 N m 2 = 1.0 ¥ 105 Pa
1 atmosphere pressure,
UNIT
SYMBOLS
meter,
kilogram,
second,
ampere,
kelvin,
PREFIXES
Factor
Prefix
Symbol
m
kg
s
A
K
mole,
hertz,
newton,
pascal,
joule,
mol
Hz
N
Pa
J
watt,
coulomb,
volt,
ohm,
henry,
W
C
V
W
H
farad,
tesla,
degree Celsius,
electron-volt,
F
T
∞C
eV
VALUES OF TRIGONOMETRIC FUNCTIONS FOR COMMON ANGLES
q
30
0
37
45
53
60
90
109
giga
G
sin q
0
12
35
2 2
45
3 2
1
106
mega
M
cosq
1
3 2
45
2 2
35
12
0
103
kilo
k
tan q
0
3 3
34
1
43
3
•
-2
centi
c
10 -3
milli
m
10 -6
micro
m
-9
nano
n
10 -12
pico
p
10
10
The following conventions are used in this exam.
I. Unless otherwise stated, the frame of reference of any problem is
assumed to be inertial.
II. The direction of any electric current is the direction of flow of positive
charge (conventional current).
III. For any isolated electric charge, the electric potential is defined as zero at
an infinite distance from the charge.
-2-
PHYSICS C: MECHANICS
SECTION I
Time—45 minutes
35 Questions
Directions: Each of the questions or incomplete statements below is followed by five suggested answers or
completions. Select the one that is best in each case and place the letter of your choice in the corresponding box on
the student answer sheet.
Note: To simplify calculations, you may use g = 10 m/ s2 in all problems.
1. Which of the following graphs of position d
versus time t corresponds to motion of an object
in a straight line with positive acceleration?
2. A ball is thrown straight up from a point 2 m
above the ground. The ball reaches a maximum
height of 3 m above its starting point and then
falls 5 m to the ground. When the ball strikes the
ground, what is its displacement from its starting
point?
(A)
(A)
(B)
(C)
(D)
(E)
Zero
8 m below
5 m below
2 m below
3 m above
(B)
3. What do acceleration and velocity have in
common?
(A) Both are scalars.
(B) Both are vectors.
(C) Both are measured in units of distance
divided by time.
(D) Both are measured in units of distance
divided by time squared.
(E) They are different names for the same
quantity.
(C)
4. Two projectiles are launched with the same initial
speed from the same location, one at a 30∞ angle
and the other at a 60∞ angle with the horizontal.
They land at the same height at which they were
launched. If air resistance is negligible, how do
the projectiles’ respective maximum heights, H30
(D)
and H60 , and times in the air, T30 and T60 ,
compare with each other?
(E)
Maximum Height
Time in Air
(A)
H30 > H60
T30 > T60
(B)
H30 > H60
T30 < T60
(C)
H30 = H60
T30 = T60
(D)
H30 < H60
T30 > T60
(E)
H30 < H60
T30 < T60
GO ON TO THE NEXT PAGE.
-3-
5. An object of mass 100 kg is initially at rest on a
horizontal frictionless surface. At time t = 0, a
horizontal force of 10 N is applied to the object
for 1 s and then removed. Which of the following
is true of the object at time t = 2 s if it is still on
the surface?
Questions 7-8
(A) It is at the same position it had at t = 0, since
a force of 10 N is not large enough to move
such a massive object.
(B) It is moving with constant nonzero
acceleration.
(C) It is moving with decreasing acceleration.
(D) It is moving at a constant speed.
(E) It has come to rest some distance away from
the position it had at t = 0.
A rock is thrown from the edge of a cliff with an
initial velocity u0 at an angle q with the horizontal
as shown above. Point P is the highest point in the
rock’s trajectory and point Q is level with the starting
point. Assume air resistance is negligible.
6. Several forces act on an object, but the object is in
equilibrium. Which of the following statements
about the object must be true?
I.
II.
III.
IV.
(A)
(B)
(C)
(D)
(E)
7. Which of the following correctly describes the
horizontal and vertical speeds and the acceleration
of the rock at point P ?
It has zero acceleration.
The net force acting on it is zero.
It is at rest.
It is moving with constant velocity.
Horizontal
Speed
(A) u0 cos q
(B)
0
(C) u0 cos q
I and II
I and III
I and IV
II and III
II and IV
Vertical
Speed
0
0
0
Acceleration
g
g
0
(D) u0 cos q
u0 sin q
g
(E)
u0 cos q
0
0
8. Which of the following correctly describes the
horizontal and vertical speeds and the acceleration
of the rock at point Q ?
Horizontal
Speed
(A) u0 cos q
(B)
0
(C) u0 cos q
Vertical
Speed
0
0
0
Acceleration
g
g
0
(D) u0 cos q
u0 sin q
g
(E)
u0 cos q
0
0
GO ON TO THE NEXT PAGE.
-4-
12. In an experiment with a simple pendulum,
measurements of the period T of the pendulum
are made for different values of its length L.
When plotted on a graph, which of the following
should result in a straight-line fit of the data?
(A) T versus L
(B) T versus L
(C) T versus L2
(D) T 2 versus L
(E) T 2 versus L2
_________________________________________
9. As shown in the figure above, a child of mass
20 kg who is running at a speed of 4.0 m/s jumps
onto a stationary sled of mass 5.0 kg on a frozen
lake. The speed at which the child and sled begin
to slide across the ice is most nearly
(A)
(B)
(C)
(D)
(E)
0.20 m/s
0.80 m/s
1.2 m/s
3.2 m/s
16 m/s
X
Sun
10. A toy spacecraft is launched directly upward.
When the toy reaches its highest point, a spring
is released and the toy splits into two parts with
masses of 0.02 kg and 0.08 kg, respectively.
Immediately after the separation, the 0.02 kg
part moves horizontally due east. Air resistance
is negligible. True statements about the 0.08 kg
part include which of the following?
Y
13. A comet moves in the Sun’s gravitational field,
following the path shown above. What happens to
its angular momentum as it moves from point X
to point Y ?
I. It could move north immediately after the
spring is released.
II. It takes longer to reach the ground than does
the 0.02 kg part.
III. It strikes the ground farther from the launch
point than does the 0.02 kg part.
(A)
(B)
(C)
(D)
(E)
(A)
(B)
(C)
(D)
It increases steadily.
It remains constant.
It decreases steadily.
It increases as it approaches the Sun and
decreases as it moves away from the Sun.
(E) It decreases as it approaches the Sun and
increases as it moves away from the Sun.
None
I only
III only
I and II only
II and III only
11. A student initially stands on a circular platform
that is free to rotate without friction about its
center. The student jumps off tangentially, setting
the platform spinning. Quantities that are
conserved for the student-platform system as the
student jumps include which of the following?
I. Angular momentum
II. Linear momentum
III. Kinetic energy
(A)
(B)
(C)
(D)
(E)
I only
II only
I and II only
II and III only
I, II, and III
GO ON TO THE NEXT PAGE.
-5-
14. Satellite X moves around Earth in a circular orbit
of radius R. Satellite Y is also in a circular orbit
around Earth, and it completes one orbit for every
eight orbits completed by satellite X. What is the
orbital radius of satellite Y ?
(A)
Questions 16-17
A toy car of mass 6 kg, moving in a straight path,
experiences a net force given by the function F = −3t.
At time t = 0, the car has a velocity of 4 m/s in the
positive direction and is located +8 m from the origin.
1
R
4
16. The car will come instantaneously to rest at time t
equal to
1
(B) R
2
(A)
(C) 2 R
(D) 4 R
(E) 8R
(B)
(C)
15. A newly discovered planet is found to have twice
the radius and five times the mass of Earth. If the
acceleration of gravity at the surface of Earth is g,
the acceleration of gravity at the surface of the
new planet is
(D)
(E)
17. Which of the following best shows a graph of
position d versus time t for the car?
2g
(A)
5
(B)
2
s
3
4
s
3
8
s
3
8s
4s
(A)
4g
5
(C) g
(D)
5g
4
(E)
5g
2
(B)
(C)
(D)
(E)
GO ON TO THE NEXT PAGE.
-6-
Questions 18-19
Questions 20-21
In the system of two blocks and a spring shown
above, blocks 1 and 2 are connected by a string that
passes over a pulley. The initially unstretched spring
connects block 1 to a rigid wall. Block 1 is released
from rest, initially slides to the right, and is eventually
brought to rest by the spring and by friction on the
horizontal surface.
A block of mass M1 on a horizontal table is
connected to a hanging block of mass M2 by a
string that passes over a pulley, as shown above.
The acceleration of the blocks is 0.6g. Assume that
friction and the mass of the string are negligible.
20. Which of the following is true of the energy of the
system during this process?
(A) The total mechanical energy of the system
is conserved.
(B) The total mechanical energy of the system
increases.
(C) The energy lost to friction is equal to the gain
in the potential energy of the spring.
(D) The potential energy lost by block 2 is less in
magnitude than the potential energy gained
by the spring.
(E) The potential energy lost by block 2 is greater
in magnitude than the potential energy
gained by the spring.
18. The tension T in the string is
(A)
(B)
(C)
(D)
(E)
zero
0.4 M2 g
0.6 M2 g
1.0 M2 g
1.6 M2 g
19. The ratio of masses M2 M1 is
(A)
(B)
(C)
(D)
(E)
0.67
1.0
1.4
1.5
1.6
21. After block 1 comes to rest, the force exerted on it
by the spring must be equal in magnitude to
(A) zero
(B) the frictional force on block 1
(C) the vector sum of the forces on block 1 due to
friction and tension in the string
(D) the sum of the weights of the two blocks
(E) the difference in the weights of the two
blocks
GO ON TO THE NEXT PAGE.
-7-
Questions 23-24
22. The graph above shows the force acting on an
object as a function of time. The change in
momentum of the object from time 0 to t is
(A) 2Ft
A moon of mass m orbits a planet of mass 49m in
an elliptical orbit as shown above. When the moon is
at point A, its distance from the center of the planet is
rA and its speed is u0 . When the moon is at point B,
(B) Ft
1
Ft
2
1
(D) Ft
4
(C)
its speed is 5u0 .
(E) zero
23. When the moon is at point A, the distance from
the moon to the center of mass of the planet-moon
system is most nearly
(A)
(B)
(C)
(D)
(E)
1
r
50 A
1
r
7 A
1
r
2 A
6
r
7 A
49
r
50 A
24. When the moon is at point B, the distance from
the moon to the center of the planet is most nearly
1
r
25 A
1
r
(B)
5 A
1
r
(C)
5 A
(A)
(D) rA
(E)
5 rA
GO ON TO THE NEXT PAGE.
-8-
Questions 25-26
The bar shown above is pivoted about one end and is initially at rest in a vertical position. The bar is displaced
slightly and as it falls it makes an angle q with the vertical at any given time, as shown above.
26. Which of the following graphs best represents the
bar’s angular velocity w as a function of time?
25. Which of the following graphs best represents the
bar’s angular acceleration a as a function of
angle q ?
(A)
(A)
(B)
(B)
(C)
(C)
(D)
(D)
(E)
(E)
GO ON TO THE NEXT PAGE.
-9-
x (m)
0
1
2
3
4
27. A stone falls from rest from the top of a building
as shown above. Which of the following graphs
best represents the stone’s angular momentum L
about the point P as a function of time?
F (N)
0
1
8
27
64
28. A specially designed spring is stretched from
equilibrium to the distances x given in the table
above, and the restoring force F is measured
and recorded in each case. What is the potential
energy of the spring when it is stretched 3 m from
equilibrium?
(A)
9
J
2
(B) 9 J
(A)
81
J
4
(D) 27 J
(C)
(B)
(E)
81
J
2
29. An object on the end of a spring with spring
constant k moves in simple harmonic motion
with amplitude A and frequency f. Which of the
following is a possible expression for the kinetic
energy of the object as a function of time t ?
(C)
(A) kA2 sin 2 (2 p ft )
(B)
(D)
1 2
kA cos2 (2 p ft )
2
1
kA sin (2 p ft )
2
(D) kA cos (2 p ft )
(C)
(E) kA (sin 2 p ft + cos2 p ft )
(E)
GO ON TO THE NEXT PAGE.
-10-
30. When a certain spring is stretched by an
Questions 31-32 refer to the following.
amount x, it produces a restoring force
of F ( x ) = - ax + bx 2 , where a and b
are constants. How much work is done by
Wall
an external force in stretching the spring
by an amount D from its equilibrium length?
(A) - aD + bD 2
Ladder
(B) a - 2bD
(C)
1 2 1 3
aD - bD
2
3
(D) - aD + bD
2
q
Floor
3
A uniform ladder of weight W leans without slipping
against a wall making an angle q with a floor as
shown above. There is friction between the ladder and
the floor, but the friction between the ladder and the
wall is negligible.
(E) -2 aD 2 + 3bD3
31. The magnitude of the normal force exerted by the
floor on the ladder is
(A) W
(B) W sin q
(C) W cos q
W
sin q
2
W
(E)
cos q
2
(D)
32. The magnitude of the friction force exerted on the
ladder by the floor is
(A) 2W tan q
(B) W
(C) W cot q
W
2
W
(E)
cot q
2
(D)
GO ON TO THE NEXT PAGE.
-11-
33. An ideal spring with spring constant k is cut in
half. What is the spring constant of either one
of the two half springs?
34. A rocket has landed on Planet X, which has half
the radius of Earth. An astronaut onboard the
rocket weighs twice as much on Planet X as on
Earth. If the escape velocity for the rocket taking
off from Earth is u0 , then its escape velocity on
Planet X is
k
2
(B) k
(A)
(A) 2u0
(C) k
(B)
(D) k 2
2u0
(C) u0
(D) u0 2
(E) 2k
(E) u0 4
GO ON TO THE NEXT PAGE.
-12-
35. Suppose that a hole is drilled through the center
of Earth to the other side along its axis. A small
object of mass m is dropped from rest into the
hole at the surface of Earth, as shown above.
If Earth is assumed to be a solid sphere of mass
M and radius R and friction is assumed to be
negligible, correct expressions for the kinetic
energy of the mass as it passes Earth’s center
include which of the following ?
1
MgR
2
1
II. mgR
2
GmM
III.
2R
I.
(A)
(B)
(C)
(D)
(E)
I only
II only
III only
I and III only
II and III only
STOP
END OF MECHANICS SECTION I
IF YOU FINISH BEFORE TIME IS CALLED,
YOU MAY CHECK YOUR WORK ON MECHANICS SECTION I ONLY.
DO NOT TURN TO ANY OTHER TEST MATERIALS.
-13-
Section II
Free-Response Questions
-14-
TABLE OF INFORMATION FOR 2008 and 2009
CONSTANTS AND CONVERSION FACTORS
Proton mass, m p = 1.67 ¥ 10 -27 kg
Electron charge magnitude,
Neutron mass, mn = 1.67 ¥ 10 -27 kg
1 electron volt, 1 eV = 1.60 ¥ 10 -19 J
Electron mass, me = 9.11 ¥ 10 -31 kg
Speed of light,
Universal gravitational
constant,
Acceleration due to gravity
at Earth’s surface,
Avogadro’s number, N 0 = 6.02 ¥ 10 23 mol-1
R = 8.31 J (moliK)
Universal gas constant,
e = 1.60 ¥ 10 -19 C
c = 3.00 ¥ 108 m s
G = 6.67 ¥ 10 -11 m 3 kgis2
g = 9.8 m s2
Boltzmann’s constant, k B = 1.38 ¥ 10 -23 J K
1 u = 1.66 ¥ 10 -27 kg = 931 MeV c 2
1 unified atomic mass unit,
h = 6.63 ¥ 10 -34 J is = 4.14 ¥ 10 -15 eV is
Planck’s constant,
hc = 1.99 ¥ 10 -25 J im = 1.24 ¥ 103 eV i nm
⑀0 = 8.85 ¥ 10 -12 C2 N im 2
Vacuum permittivity,
Coulomb’s law constant, k = 1 4 p⑀0 = 9.0 ¥ 109 N im 2 C2
m0 = 4 p ¥ 10 -7 (T im) A
Vacuum permeability,
Magnetic constant, k ¢ = m0 4 p = 10 -7 (T i m) A
1 atm = 1.0 ¥ 105 N m 2 = 1.0 ¥ 105 Pa
1 atmosphere pressure,
UNIT
SYMBOLS
meter,
kilogram,
second,
ampere,
kelvin,
PREFIXES
m
kg
s
A
K
mole,
hertz,
newton,
pascal,
joule,
mol
Hz
N
Pa
J
watt,
coulomb,
volt,
ohm,
henry,
W
C
V
W
H
farad,
tesla,
degree Celsius,
electron-volt,
F
T
∞C
eV
VALUES OF TRIGONOMETRIC FUNCTIONS FOR COMMON ANGLES
Prefix
Symbol
q
0
30
37
45
53
60
giga
G
sin q
0
12
35
2 2
45
3 2
1
106
mega
M
cosq
1
3 2
45
2 2
35
12
0
103
kilo
k
tan q
0
3 3
34
1
43
3
•
10 -2
centi
c
10 -3
milli
m
10 -6
micro
m
10 -9
nano
n
10 -12
pico
p
Factor
10
9
90
The following conventions are used in this exam.
I. Unless otherwise stated, the frame of reference of any problem is
assumed to be inertial.
II. The direction of any electric current is the direction of flow of positive
charge (conventional current).
III. For any isolated electric charge, the electric potential is defined as zero at
an infinite distance from the charge.
-15-
ADVANCED PLACEMENT PHYSICS C EQUATIONS FOR 2008 and 2009
MECHANICS
u = u0 + at
1 2
at
2
x = x0 + u0 t +
u 2 = u0 2 + 2 a ( x - x0 )
 F = Fnet = ma
F=
dp
dt
J = Ú F dt = Dp
p = mv
Ffric £ m N
W =
ÚF
K =
1
mu 2
2
P =
dW
dt
∑
dr
P = Fⴢv
DUg = mgh
ac =
u2
= w2r
r
a
F
f
h
I
J
K
k
=
=
=
=
=
=
=
=
=
L =
m=
N =
P =
p =
r =
r =
T =
t =
U=
u =
W=
x =
m=
q =
t =
w=
a=
ELECTRICITY AND MAGNETISM
acceleration
force
frequency
height
rotational inertia
impulse
kinetic energy
spring constant
length
angular momentum
mass
normal force
power
momentum
radius or distance
position vector
period
time
potential energy
velocity or speed
work done on a system
position
coefficient of friction
angle
torque
angular speed
angular acceleration
Fs = - kx
t=r¥F
 t = t net = I a
Us =
1 2
kx
2
2p
1
=
f
w
I = Ú r 2 dm = Â mr 2
T =
rcm =  mr  m
Ts = 2 p
m
k
u = rw
L = r ¥ p = Iw
Tp = 2 p
1 2
Iw
2
FG = -
Gm1m2
UG = -
Gm1m2
r
K =
w = w0 + a t
q = q0 + w0 t +
1 2
at
2
1 q1q2
4 p⑀0 r 2
E=
F
q
ÚE
UE
rˆ
⑀0
dV
dr
E = -
q
 rii
1
4 p⑀0
V =
Q
dA =
∑
i
1 q1q2
= qV =
4 p⑀0 r
C =
Q
V
C =
k ⑀0 A
d
Cp =
 Ci
i
1
1
=Â
Cs
i Ci
A
B
C
d
E
=
=
=
=
=
e=
F =
I =
J =
L =
=
n =
area
magnetic field
capacitance
distance
electric field
emf
force
current
current density
inductance
length
number of loops of wire
per unit length
number of charge carriers
per unit volume
power
charge
point charge
resistance
distance
time
potential or stored energy
electric potential
velocity or speed
resistivity
N =
P =
Q=
q =
R =
r =
t =
U=
V=
u =
r=
fm = magnetic flux
k = dielectric constant
dQ
I =
dt
Uc =
1
1
QV = CV 2
2
2
R=
r
A
ÚB
∑
d ᐉ = m0 I
m0 I d ᐉ ¥ r
4p r 3
dB =
Ú I dᐉ ¥ B
E = rJ
F=
I = Neud A
Bs = m0 nI
V = IR
fm =
Rs =
g
r2
F =
1
=
Rp
 Ri
i
i
i
P = IV
FM = qv ¥ B
-16-
=-
e
= -L
UL =
∑
dA
d fm
dt
e
1
ÂR
ÚB
dI
dt
1 2
LI
2
ADVANCED PLACEMENT PHYSICS C EQUATIONS FOR 2008 and 2009
GEOMETRY AND TRIGONOMETRY
Rectangle
A = bh
Triangle
A=
A=
C=
V=
S =
b =
h =
=
w=
r =
1
bh
2
Circle
A = pr 2
C = 2pr
Parallelepiped
V = wh
Cylinder
CALCULUS
area
circumference
volume
surface area
base
height
length
width
radius
df
d f du
=
dx
du dx
d n
( x ) = nx n -1
dx
d x
(e ) = e x
dx
d
(1n x ) = 1
dx
x
d
(sin x ) = cos x
dx
d
(cos x ) = - sin x
dx
V = pr 2
S = 2pr + 2 pr
2
Sphere
V =
4 3
pr
3
tan q =
a
b
Úe
x
dx = e x
dx
= ln x
x
Ú sin x dx = - cos x
a
c
b
c
dx =
Ú cos x dx = sin x
a 2 + b2 = c 2
cos q =
n
Ú
S = 4pr 2
Right Triangle
sin q =
1
x n + 1 , n π -1
n +1
Úx
c
a
90°
q
b
-17-
PHYSICS C: MECHANICS
SECTION II
Time—45 minutes
3 Questions
Directions: Answer all three questions. The suggested time is about 15 minutes for answering each of the questions,
which are worth 15 points each. The parts within a question may not have equal weight. All final numerical answers
should include appropriate units. Credit depends on the quality of your solutions and explanations, so you should
show your work. Credit also depends on demonstrating that you know which physical principles would be
appropriate to apply in a particular situation. Therefore, you should clearly indicate which part of a question your
work is for.
Mech. 1.
A chunk of clay of mass 0.20 kg is thrown from the ground with an initial speed of 12 m s at an angle of 30∞
with the horizontal, as shown above. At the top of its trajectory, the clay strikes a small block of mass 2.3 kg
suspended from a 3.0 m long string. The clay sticks to the block, which then swings freely. Neglect air resistance.
(a)
Calculate the horizontal distance D between the launching point of the clay and a point on the floor directly
below the initial position of the block.
(b)
Calculate the speed of the block-clay system immediately after the collision with the clay.
(c)
Calculate the angle q through which the block-clay system will rise before coming momentarily to rest.
(d)
Calculate the time between when the block is struck and when it first returns to its original position.
(e)
The procedure is repeated with a chunk of clay of greater mass. Indicate whether the new angle q will be
greater than, less than, or the same as that determined in (c).
___ Greater
___ Less
___ The same
Justify your answer.
GO ON TO THE NEXT PAGE.
-18-
Mech. 2.
In the lab apparatus above, a force sensor attached to a cart is connected by a string to a block. The string passes
over a pulley. The block is allowed to fall, accelerating the cart. A computer attached to the force sensor and a
motion detector displays the position, the speed, and the force applied to the cart at five different locations, as
given in the table below. The square of the speed is also provided.
(
Position (m)
Speed ( m s )
Force (N)
Speed 2 m 2 s2
0.00
0.10
0.20
0.30
0.40
0.55
0.66
0.85
0.94
1.08
0.84
0.85
0.84
0.83
0.85
0.30
0.44
0.73
0.88
1.17
)
(a) i. Determine the average force exerted by the string.
ii. Estimate the work done by the average force on the cart during the time that data was taken.
(b) i. On the axes below, graph the square of the speed versus position.
ii. Draw a best-fit line through the points.
GO ON TO THE NEXT PAGE.
-19-
(c) Calculate the acceleration of the cart from the best-fit line.
The mass of the cart is 0.65 kg.
(d) Use a method different from that used in (a) ii. to calculate the work done on the cart from the data given.
(e) Indicate whether the values you obtained in (a) and (d) are in agreement. If they are, explain why they
should be. If they are not, indicate a possible cause of the discrepancy.
GO ON TO THE NEXT PAGE.
-20-
Mech. 3.
A student holds one end of a thread, which is wrapped around a cylindrical spool, as shown above. The student
then drops the spool from a height h above the floor, and the thread unwinds as it falls. The spool has a mass M
and a radius R, and the thread has negligible mass. The spool can be approximated as a solid cylinder of moment
1
of inertia I = MR 2 . Express your answers in terms of M, R, h, and fundamental constants.
2
(a) Calculate the linear acceleration of the spool as it falls.
(b) Calculate the angular velocity of the spool just before it strikes the floor.
At time t = 0, the spinning spool lands on the floor without bouncing and comes free from the thread. It continues
to spin, but slips on the floor’s surface while doing so. Assume a constant coefficient of sliding friction m .
(c) Calculate the angular velocity of the spool as a function of time t.
(d) Calculate the horizontal speed of the spool as a function of time, assuming the horizontal speed is zero at
time t = 0.
(e) At what time does slipping between the spool and floor cease?
STOP
END OF EXAM
-21-
Name: ____________________________________
AP® Physics C: Mechanics
Student Answer Sheet for Multiple-Choice Section
No.
1
Answer
No.
31
2
32
3
33
4
34
5
35
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
-22-
Answer
AP® Physics C: Mechanics
Multiple-Choice Answer Key
No.
1
Correct
Answer
D
No.
31
Correct
Answer
A
2
D
32
E
3
B
33
E
4
E
34
C
5
D
35
E
6
A
7
A
8
D
9
D
10
A
11
A
12
D
13
B
14
D
15
D
16
E
17
E
18
B
19
D
20
E
21
C
22
C
23
E
24
B
25
D
26
E
27
C
28
C
29
B
30
C
-23-
AP® Physics C: Mechanics
Free-Response Scoring Guidelines
General Notes about AP Physics Practice Exam Scoring Guidelines
1.
The solutions contain a common method of solving the free-response questions and the allocation of
points for the solutions. Some also contain a common alternate solution. They are typical of draft
guidelines developed before student solutions are available. Teachers should feel free to make
modifications based on their students’ responses.
2. The scoring guidelines typically show numerical results using the value g = 9.8 m s2 , but use of
10 m s2 is of course also acceptable. Solutions usually show numerical answers using both values
when they are significantly different.
The following rules apply to the official scoring of AP Physics Exams.
3.
All correct methods of solution receive appropriate credit for correct work.
4. Generally, double penalty for errors is avoided. For example, if an incorrect answer to part (a) is
correctly substituted into an otherwise correct solution to part (b), full credit will usually be awarded.
One exception to this may be cases when the numerical answer to a later part should be easily
recognized as wrong, e.g., a speed faster than the speed of light in vacuum.
5. Implicit statements of concepts normally receive credit. For example, if use of the equation expressing a
particular concept is worth 1 point, and a student’s solution contains the application of that equation to
the problem but the student does not write the basic equation, the point is still awarded. However, when
students are asked to derive an expression, it is normally expected that they will begin by writing one or
more fundamental equations, such as those given on the AP Physics Exam equation sheet. For a
description of the use of such terms as “derive” and “calculate” on the exams, and what is expected for
each, see “The Free-Response Sections⎯Student Presentation” in the AP Physics Course Description.
6.
Strict rules regarding significant digits are usually not applied to numerical answers. However, in some
cases answers containing too many digits may be penalized. In general, two to four significant digits are
acceptable. Numerical answers that differ from the published answer due to differences in rounding
throughout the question typically receive full credit. Exceptions to these guidelines usually occur when
rounding makes a difference in obtaining a reasonable answer. For example, suppose a solution requires
subtracting two numbers that should have five significant figures and that differ starting with the fourth
digit (e.g., 20.295 and 20.278). Rounding to three digits will lose the accuracy required to determine the
difference in the numbers, and some credit may be lost.
-24-
AP® Physics C: Mechanics
Free-Response Scoring Guidelines
Question 1
15 points total
(a)
Distribution
of points
5 points
For using a correct kinematic equation to determine the time of flight
u y = uy 0 - gt
1 point
For correctly using the vertical component of the initial velocity
u y 0 = u0 sin30D
1 point
0 = u0 sin 30∞ - gt
For a correct expression for the time
u0 sin30D
t =
g
1 point
(12.0 m/s) sin 30D
= 0.61 s
9.8 m s2
For a correct kinematic equation including correct use of the horizontal component of the
initial velocity
D = ux t
1 point
t =
u x = u0 cos 30D = 10.4 m/s
D = (10.4 m/s) (0.61 s)
For the correct answer
1 point
D = 6.4 m (or 6.2 m using g = 10 m s )
2
(b)
3 points
For applying conservation of momentum
For correctly using the sum of the clay and block masses after the collision
mu x = (m + M ) u f
m
u
m+M x
0.20 kg
uf =
(10.4 m/s)
0.20 kg + 2.3 kg
For the correct answer
u f = 0.83 m/s
1 point
1 point
uf =
1 point
-25-
AP® Physics C: Mechanics
Free-Response Scoring Guidelines
Question 1 (continued)
Distribution
of points
(c)
3 points
For applying conservation of energy
1
(m + M ) u 2f = (m + M ) gDh
2
Solving for Dh
1 point
u 2f
(0.83 m s)2
Dh =
=
= 0.035 m
2g
2 9.8 m s2
(
)
For applying the correct trigonometry relating Dh and the length of the string A
Dh = A - A cos q
A - Dh
3.0 - 0.035
= 8.8∞
q = cos -1
= cos-1
A
3.0
For the correct answer
(
q = 8.8∞
(d)
)
(
1 point
)
1 point
(or 8.7∞ using g = 10 m s )
2
2 points
For correctly calculating the period
A
T = 2p
g
1 point
3.0 m
= 3.5 s
9.8 m s2
For an indication that the requested time is one-half the period
1
t = T = 1.7 s
2
T = 2p
(e)
1 point
2 points
For correctly indicating that the new angle is greater
For a correct justification
For example: Substituting the expression for u f from part (b) into the first expression for
(
)
2
1
m
u x2
2g m + M
Increasing the mass m increases the height achieved by the pendulum; hence, the
resulting angle is greater than that achieved with the original mass.
Dh from part (c): Dh =
-26-
1 point
1 point
AP® Physics C: Mechanics
Free-Response Scoring Guidelines
Question 2
15 points total
Distribution
of points
(a)
(i)
1 point
For a correct determination of the average force
(0.84 + 0.85 + 0.84 + 0.83 + 0.85) N
Favg =
= 0.84 N
5
(ii)
1 point
1 point
W = Favg x
W = (0.84 N )(0.40 m )
For the correct answer
W = 0.34 J
1 point
(b)
(i)
2 points
For correctly plotting four of the five given points
(ii)
2 points
1 point
For correctly drawing a straight line that has at least two data points above the line and two
points below the line
-27-
1 point
AP® Physics C: Mechanics
Free-Response Scoring Guidelines
Question 2 (continued)
Distribution
of points
(c)
4 points
For applying the correct kinematic equation to determine the acceleration
u 2 = u02 + 2ax
For correctly equating the slope of the graph to twice the acceleration
Slope = 2a
For correct determination of the slope using points clearly on the best-fit line
(Using data points not on the line does not receive credit.)
For example, using points on the graph shown:
(0.93 - 0.49) m 2 s2
Slope =
= 2.20 m s2
(0.3 - 0.1) m
1 point
1 point
For a numerical answer in the range 0.83 to 1.3 m s2
1 point
1 point
a = 1.10 m s2
(d)
3 points
For correct identification of a valid alternative method
For example, conservation of energy:
1
W = m u 2f - ui2
2
1
2
2
W = ( 0.65 kg ) (1.08 m s) - ( 0.55 m s)
2
For the correct answer
W = 0.28 J
(
)
(
(e)
)
1 point
2 points
For a reasonable answer based on the two values of work obtained
For a correct and substantive explanation
For example: The work calculated in (d) is less than the work calculated in (a). Part (a)
includes only the positive work performed on the cart by the string. There could be
energy dissipated due to friction, which is included when you calculate the actual
change in kinetic energy in part (d).
Units
2 points
1 point
1 point
1 point
For correct units in all final numerical answers
1 point
-28-
AP® Physics C: Mechanics
Free-Response Scoring Guidelines
Question 3
15 points total
(a)
Distribution
of points
4 points
For correctly applying Newton’s second law
Ma = Mg - T
For correctly applying a torque equation relating T and a
t = TR = I a
For correctly incorporating the relationship between a and a
a
TR = I a = I
R
a
T = I 2
R
Substituting into the equation for Newton’s second law
Ia
Ma = Mg - 2
R
I
Mg = a ÊÁ M + 2 ˆ˜
Ë
R ¯
Substituting the given expression for I
1 1
Mg = a ÊÁ M + 2
MR 2 ˆ˜
Ë
¯
2
R
M
3
Mg = a M +
= Ma
2
2
For the correct answer
2
a= g
3
(
(b)
)
(
1 point
1 point
1 point
)
1 point
3 points
For applying conservation of energy
For correctly including all three terms (gravitational potential, linear kinetic, rotational
kinetic)
1
1
Mgh = M u 2 + I w 2
2
2
Substituting u = r w and the given expression for I
1
1 1
Mgh = MR 2 w 2 +
MR 2 w 2
2
2 2
3
2 2
Mgh = MR w
4
For the correct answer
4 gh
w=
3R2
(
1 point
1 point
)
-29-
1 point
AP® Physics C: Mechanics
Free-Response Scoring Guidelines
Question 3 (continued)
Alternate solution
For correctly substituting the acceleration from part (a) into the kinematic equation
u 2 = u02 + 2a ( y - y0 )
u2 = 2
Distribution
of points
Alternate points
1 point
( 23 g) h = 43gh
For substituting u into the expression w = u / R
4 gh 3
u
w =
=
R
R
For the correct answer
4 gh
w=
3R2
Note: Another alternate solution, obtained by combining the time determined by kinematics
with the expression relating angular impulse and change in angular momentum, would
also be acceptable if correctly implemented.
(c)
1 point
4 points
For applying the torque equation
For a correct expression for the torque due to friction
t = I a = - m MgR
1
MR 2 a = - m MgR
2
2 mg
a =R
For applying the appropriate rotational kinematic equation
w = w0 + a t
For using the result of part (b) as the initial angular velocity
4 gh 2 mg
t
w=
R
3R 2
(d)
1 point
1 point
1 point
1 point
1 point
2 points
For correct use of Newton’s second law and a kinematic equation
Ma = m Mg
a = mg
u = at
For the correct answer
u = mgt
-30-
1 point
1 point
AP® Physics C: Mechanics
Free-Response Scoring Guidelines
Question 3 (continued)
Distribution
of points
(e)
2 points
For correctly stating the condition for cessation of slipping
u = Rw
For correctly substituting results from parts (c) and (d)
2 mg ˘
È 4 gh
m gt = R Í
t
2
R ˙˚
Î 3R
gh
m gt = 2
- 2 m gt
3
2 h
4h
t =
=
3 m 3g
27 m 2 g
-31-
1 point
1 point